Question

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 potential of the cell 4. Increasing the mass of the Cu will change the initial potential of the cell 5. Cu2+ is being reduced during the reaction

Answer 1

Cu²⁺⁺   + 2e^- ----> Cu ; 0.34V

Fe³⁺  + 3e^- -----> Fe ; -0.0036V

We must combine these half-reactions to get a positive cell potential. we reverse the Fe half-reaction

Fe(s) → Fe²⁺(aq) + 2e⁻; E₀ =-0.0036V

2×[Cu⁺(aq) + e⁻ → Cu(s)]; E₀ = 0.34 V

Fe(s) + 2Cu⁺(aq) → Fe²⁺(aq) + 2Cu(s); E₀ = + 0.3364 V

This tells us that Fe is spontaneously oxidized to Fe²⁺(at the anode)

The solid Fe anode gradually disappears as it is converted to the soluble Fe²⁺ ions. The mass of the anode decreases during the cell operation

The Cu⁺ ions are spontaneously reduced to solid copper (at the cathode). They stick to the cathode, so the mass of the cathode increases during the cell operation.

In any galvanic cell with reactive electrodes, the mass of the cathode increases and the mass of the anode decreases while the cell is operating.

E⁰cell = E⁰red + E⁰ox

Ecell = E⁰cell − ( RT/nF)lnQ ; n= number of moles of electrons transferred in the balanced equation  

1) if concentration of copper decreased then the overall cell potential will decrease according to above equation of E⁰_cell (TRUE )

2)FALSE , since iron is oxidised , it will contribute negative potential to the overall cell potential

3) FALSE , we have both H+ and OH- here , contribute equal amount of cell potential but opposite so no effect on cell potential .

4) TRUE

5)TRUE