Calculate the E_ cell for Zn_ (s) | Zn^ ++ _ (0.1M) || Cn^ ++ _ (0.1M) | Cr_ (s) at 25^0C E^0_ cell is 0.02 V

MHT CET 2023 Chemistry question

Calculate the for Zn_{(s)} | Zn^{++}_{(0.1M)} || Cn^{++}_{(0.1M)} | Cr_{(s)} at is 0.02 V

(A)-0.05 V
(B)0.03VCorrect
(C)-0.06 V
(D)0.07 V

Answer: (B)

Solution

The cell has a zinc anode and a chromium cathode, so the intended reaction is 3Zn + 2Cr^{3+} \rightarrow 3Zn^{2+} + 2Cr with electrons transferred (the printed cathode ion is a typo for ). Nernst equation at 298 K: , where \frac{[Zn^{2+}]^3}{[Cr^{3+}]^2}$ =

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Question ID VK-097261