Question

Part A. When the Cu²⁺ concentration is 1.24 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 2.820V. What is the Mg²⁺ concentration?

Cu²⁺(aq) + Mg(s) ----> Cu(s) + Mg²⁺(aq)

Answer: ________ M

Part B. When the Hg²⁺ concentration is 1.13 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.718V. What is the Zn²⁺ concentration?

Hg²⁺(aq) + Zn(s) -----> Hg(l) + Zn²⁺(aq)

Answer: _______ M

Answer 1

Given:-

Part A.

molar concentration of [Cu²⁺] = 1.24 M

cell potential (E_cell) = 2.820 V.

Temperature(T) = 298 K

molar concentration of [Mg²⁺] = ?

As we know that
Cu²⁺_(aq) + Mg_(s)    Cu_(s)   + Mg²⁺_(aq)

First half cell reaction at the anode:-

  Mg_(s)       Mg²⁺_(aq) + 2e^- (oxidation) ;   E⁰_Mg/Mg2+ = - 2.37 V ----------------------- (1)

Second half cell reaction at the cathode:-

Cu²⁺_(aq) + 2e^-    Cu_(s) (reduction) ;   E⁰_Cu2+/Cu = 0.34 V --------------------------(2)

from equation no. 1 and 2 we get

Mg_(s)       Mg²⁺_(aq) + 2e^-

Cu²⁺_(aq) + 2e^-    Cu_(s)

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Cu²⁺_(aq) + Mg_(s)    Cu_(s)   + Mg²⁺_(aq) (cell reaction)

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above equation shows that total no. of electron involves in cell reaction (n) = 2

As we know that according to the Nernst equation

Cell potential (E_cell) = E⁰_Cell - 2.303RT/ nF log [Mg²⁺] / [Cu²⁺]

E_cell = ( E⁰_cathode - E⁰_anode) - 2.303RT/ nF log [Mg²⁺] / [Cu²⁺]

E_cell = ( E⁰_Cu2+/Cu - E⁰_Mg/Mg2+) - 2.303RT/ nF log [Mg²⁺] / [Cu²⁺]

2.820V = [ (0.34 V) - ( - 2.37 V)]- 2.303 8.314 J/K-mol 298 K / 2 96500 log [Mg²⁺] / [1.24 M]

2.820V = (0.34 V + 2.37 V)- 0.029563 J/mol (log[Mg²⁺] -log[1.24] )

2.820V = (2.71 V - 0.029563 V) (log[Mg²⁺] - 0.09342 )

2.820V =  2.6804 V (log[Mg²⁺] - 0.09342 )

(log[Mg²⁺] - 0.09342 ) = 2.820 V / 2.6804 V

log[Mg²⁺] - 0.09342 = 1.05208

log[Mg²⁺] = 1.05208 + 0.09342

log[Mg²⁺] = 1.1455

[Mg²⁺] = 13.979769 M ( i.e the answer)

Part B.

molar concentration of [Hg²⁺] = 1.13 M

cell potential (E_cell) = 1.718 V

Temperature(T) = 298 K

molar concentration of [Zn²⁺] = ?

As we know that
Hg²⁺_(aq) + Zn_(s)    Hg_(l)   + Zn²⁺_(aq)

First half cell reaction at the anode:-

  Zn_(s)       Zn²⁺_(aq) + 2e^- (oxidation) ;   E⁰_{Zn /Zn2+} = - 0.76 V -------------------- (1)

Second half cell reaction at the cathode:-

Hg²⁺_(aq) + 2e^-    Hg_(l) (reduction) ;   E⁰_{Hg2+ /Hg} = 0.85 V -------------------------(2)

from equation no. 1 and 2 we get

Zn_(s)       Zn²⁺_(aq) + 2e^-  

Hg²⁺_(aq) + 2e^-    Hg_(l)

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Hg²⁺_(aq) + Zn_(s)    Hg_(l)   + Zn²⁺_(aq) (cell reaction)

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above equation shows that total no. of electron involves in cell reaction (n) = 2

As we know that according to the Nernst equation

Cell potential (E_cell) = E⁰_Cell - 2.303RT/ nF log [Zn²⁺] / [Hg²⁺]

E_cell = ( E⁰_cathode - E⁰_anode) - 2.303RT/ nF log [Zn²⁺] / [Hg²⁺]

E_cell = ( E⁰_{Hg2+ /Hg}  - E⁰_{Zn /Zn2+}) - 2.303RT/ nF log [Zn²⁺] / [Hg²⁺]

1.718 V =  [ (0.85 V) - ( - 0.76 V)]- 2.303 8.314 J/K-mol 298 K / 2 96500 log [Zn²⁺] / [1.13 M]

1.718 V = (0.85 V + 0.76 V)- 0.029563 J/mol (log[Zn²⁺] -log[1.13] )

1.718 V =(1.61 V - 0.029563 V) (log[Zn²⁺] - 0.05308 )

1.718 V =  1.5804 V (log[Zn²⁺] - 0.05308 )

(log[Zn²⁺] - 0.05308 ) = 1.718 V / 1.5804 V

log[Zn²⁺] - 0.05308 = 1.08707

log[Zn²⁺] = 1.08707 + 0.05308

log[Zn²⁺] = 1.14015

[Zn²⁺] = 13.808611 M ( i.e the answer)