JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Redox Reactions and Electrochemistry

For a cell,

Cu (s) ∣ Cu^{2 +} (0.001 M) ∣∣ Ag^{+} (0.01 M) ∣ Ag (s)

the cell potential is found to be 0.43 V at 298 K. The magnitude of standard electrode potential for

Cu^{2 +} / Cu

is ______

\times 1 0^{- 2} V

[

Given :

E_{Ag^{+} / Ag}^{Θ} = 0.80 V

and

\frac{2.303 RT}{F} = 0.06 V]

Answer: 34

Solution

Cell reaction:

Cu + 2 Ag X^{+} Cu X^{2 +} + 2 Ag

n = 2

Using the Nernst equation:

E_{cell} = E_{cell}^{\circ} - \frac{0.06}{2} lo g \frac{[ Cu X ^{2 +} ]}{[ Ag X ^{+} ] ^{2}}

0.43 = E_{cell}^{\circ} - 0.03 lo g \frac{0.001}{( 0.01 ) ^{2}}

0.43 = E_{cell}^{\circ} - 0.03 lo g \frac{0.001}{0.0001}

0.43 = E_{cell}^{\circ} - 0.03 lo g 10

0.43 = E_{cell}^{\circ} - 0.03

E_{cell}^{\circ} = 0.46

V

Since

E_{cell}^{\circ} = E_{Ag X^{+} / Ag}^{\circ} - E_{Cu X^{2 +} / Cu}^{\circ}

0.46 = 0.80 - E_{Cu X^{2 +} / Cu}^{\circ}

E_{Cu X^{2 +} / Cu}^{\circ} = 0.34

V =

34 \times 1 0^{- 2}

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer