Given the equilibrium constant: KC of the reaction : Cu (s) + 2 Ag^+ (aq) \to Cu^2+ (aq) + 2 Ag (s) is 10 \times 1015, calculate the E _cell^0 of this reaciton at 298 K [2.303 RT \over F at 298 K = 0.059V]

Electrochemistry PYQ — JEE Main 2019 · 11 Jan · Shift 2 · Q11 with Solution

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Correct answer: C

The reaction is:

Cu(s) + 2Ag^+(aq) \to Cu^{2+}(aq) + 2Ag(s)

Number of electrons transferred:

So,

n = 2

At $298\,K$ ,

E^\circ_{cell} = \frac{0.059}{n}\log K

Given:

K = 10 \times 10^{15} = 10^{16}

Thus,

\log K = \log(10^{16}) = 16

E^\circ_{cell} = \frac{0.059}{2} \times 16

E^\circ_{cell} = 0.0295 \times 16 = 0.472 \text{ V}

The calculated value is approximately:

E^\circ_{cell} \approx 0.47\text{ V}

Among the given options, the closest is:

\boxed{0.4736\text{ V}}

So, the correct option is C.