Question

In: Chemistry

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80...

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Adding equal amounts of water to both half reactions.

Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Adding Ag+ ions to the silver half reaction (assume no volume change)

Solutions

Expert Solution

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q   

where                              Ecell = actual cell potential

                                       E°cell­ ­​ = standard cell potential

                                       R = the universal gas constant = 8.314472(15) J K−1 mol−1

                                       T = the temperature in kelvins

                                       n = the number of moles of electrons transferred                                     

                                       F = the Faraday constant, the number of coulombs per mole of electrons:

                                               (F = 9.64853399(24)×104 C mol−1)

                                      Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from                             the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.


Related Solutions

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80...
Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Pb2+ + 2e- → Pb -0.13 V How would the following changes alter the potential of the cell? Adding Ag+ ions to the silver half reaction (assume no volume change) Adding Pb2+ ions to the lead half reaction (assuming no volume change). Removing Pb2+ ions from solution by precipitating them out of the lead half reaction (assume no volume change). Adding equal amounts...
    Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag;...
    Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag; 0.80 V Cu2+ + 2 e- → Cu; 0.34 V Which of the following changes will decrease the potential of the cell? Adding Cu2+ ions to the copper half reaction (assuming no volume change). Adding equal amounts of water to both half reactions. Adding Ag+ ions to the silver half reaction (assume no volume change) Removing Cu2+ ions from solution by precipitating them out...
1. A galvanic cell is based on the following half-reactions at 285 K: Ag+ + e-...
1. A galvanic cell is based on the following half-reactions at 285 K: Ag+ + e- → Ag     Eo = 0.803 V   H2O2 (aq) + 2 H+ + 2 e- → 2 H2O     Eo = 1.78 V What will the potential of this cell be when [Ag+] = 0.571 M, [H+] = 0.00341 M, and [H2O2] = 0.895 M? Please show full work .
Consider a galvanic cell based on the following half reactions: E° (V) Al3+ + 3e- →...
Consider a galvanic cell based on the following half reactions: E° (V) Al3+ + 3e- → Al -1.66 Ni2+ + 2e- → Ni -0.23 If this cell is set up at 25°C with [Ni2+] = 1.00 × 10-3M and [Al3+] = 2.00 × 10-2M, the expected cell potential is what?
Consider a galvanic cell based upon the following half reactions: Cu2+ + 2e- → Cu 0.34...
Consider a galvanic cell based upon the following half reactions: Cu2+ + 2e- → Cu 0.34 V Fe3+ + 3e- → Fe -0.0036 V How many of the following responses are true? 1. Decreasing the concentration of Cu2+ (assuming no volume change) will decrease the potential of the cell 2. Increasing the concentration of Fe3+ (assuming no volume change) will increase the potential of the cell 3. Adding equal amounts of water to both half reaction vessels will increase the...
Consider a galvanic cell based on the following two half reactions at 298 K: SRP Fe3+...
Consider a galvanic cell based on the following two half reactions at 298 K: SRP Fe3+ + e-  Fe2+ 0.77 V Cu2+ + 2 e-  Cu 0.34 V How many of the following changes will increase the potential of the cell? Why? 1. Increasing the concentration of Fe+3 ions 2. Increasing the concentration of Cu2+ ions 3. Removing equal volumes of water in both half reactions through evaporation 4. Increasing the concentration of Fe2+ ions 5. Adding a...
A concentration cell based on the following half reaction at 309 K Ag+ + e- →...
A concentration cell based on the following half reaction at 309 K Ag+ + e- → Ag SRP = 0.80 V has initial concentrations of 1.37 M Ag+, 0.269 M Ag+, and a potential of 0.04334 V at these conditions. After 9.3 hours, the new potential of the cell is found to be 0.01406 V. What is the concentration of Ag+ at the cathode at this new potential?
A concentration cell based on the following half reaction at 283 K Ag+ + e- →...
A concentration cell based on the following half reaction at 283 K Ag+ + e- → Ag       SRP = 0.80 V has initial concentrations of 1.35 M Ag+, 0.407 M Ag+, and a potential of 0.02924 V at these conditions. After 3.4 hours, the new potential of the cell is found to be 0.01157 V. What is the concentration of Ag+ at the cathode at this new potential?
A concentration cell based on the following half reaction at 312 k ag+ + e- -------->...
A concentration cell based on the following half reaction at 312 k ag+ + e- --------> ag srp = 0.80 v has initial concentrations of 1.25 m ag+, 0.221 m ag+, and a potential of 0.04865 v at these conditions. after 8.3 hours, the new potential of the cell is found to be 0.01323 v. what is the concentration of ag+ at the cathode at this new potential? Please explain all the steps used to find the answer.
A. A standard galvanic cell is constructed with Cr3+|Cr and Ag+|Ag half cell compartments connected by...
A. A standard galvanic cell is constructed with Cr3+|Cr and Ag+|Ag half cell compartments connected by a salt bridge. Which of the following statements are correct? Hint: Refer to a table of standard reduction potentials. (Choose all that apply.) 1.As the cell runs, anions will migrate from the Cr3+|Cr compartment to the Ag+|Ag compartment. 2.The anode compartment is the Ag+|Ag compartment. 3.Ag is oxidized at the anode. 4.Cr is oxidized at the anode. 5.In the external circuit, electrons flow from...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT