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← JEE Main 2022 · 25 Jul · Shift 1
ChemistryInteger+4 / -1Question 84 of 90

JEE Main 2022 · 25 Jul · Shift 1, Chemistry question 84

The cell potential for Zn∣Zn2+(aq)∣∣Snx+∣Sn\mathrm{Zn|Zn^{2+}(aq)||Sn^{x+}|Sn} is 0.801 V at 298 K. The reaction quotient for the above reaction is 10−210^{-2}. The number of electrons involved in the given electrochemical cell reaction is ______.

(Given : EZn2+∣Zn∘=−0.763E^\circ_{\mathrm{Zn^{2+}|Zn}}=-0.763 V, ESnx+∣Sn∘=+0.008E^\circ_{\mathrm{Sn^{x+}|Sn}}=+0.008 V and 2.303RTF=0.06\dfrac{2.303RT}F=0.06 V)

Answer and worked solution

Official answer: 4

Nernst. 0.801=0.771−0.06nlog⁡10−2=0.771+0.12n0.801=0.771-\frac{0.06}n\log10^{-2}=0.771+\frac{0.12}n, so n=4n=4.

Answer: 4