For the galvanic cell, $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.02 \mathrm{M}) \rightarrow \mathrm{Zn}^{2+}(0.04 \mathrm{M})+\mathrm{Cu}(\mathrm{s})$, $\mathrm{E}_{\text {cell }}=$ $\_\_\_\_$ $\times 10^{-2} \mathrm{~V}$. (Nearest integer) [Use: $\mathrm{E}^{0}_{ \mathrm{Cu} / \mathrm{Cu}^{2+}}=-0.34 \mathrm{~V}, \mathrm{E}^{0}_{ \mathrm{Zn} / \mathrm{Zn}^{2+}}=+0.76 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$]

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Redox Reactions and Electrochemistry

For the galvanic cell,

Zn (s) + Cu^{2 +} (0.02 M) \to Zn^{2 +} (0.04 M) + Cu (s)

E_{cell} =

____

\times 1 0^{- 2} V

.
(Nearest integer)
[Use:

E_{Cu / Cu^{2 +}}^{0} = - 0.34 V, E_{Zn / Zn^{2 +}}^{0} = + 0.76 V, \frac{2.303 RT}{F} = 0.059 V

]

Answer: 109

The potentials supplied are written as oxidation potentials, so the corresponding reduction potentials are

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

V and

E_{Z n^{2 +} /Zn}^{\circ} = - 0.76

V, giving

E_{cell}^{\circ} = 0.34 - (- 0.76) = 1.10 V

Two electrons are transferred, so the Nernst equation gives

E_{cell} = 1.10 - \frac{0.059}{2} lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]} = 1.10 - \frac{0.059}{2} lo g \frac{0.04}{0.02}

E_{cell} = 1.10 - 0.0295 \times 0.301 = 1.091 V

The cell potential is

109 \times 1 0^{- 2}

V to the nearest integer.

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Integer answer