A) spontaneous
B) non-spontaneous
C) equilibrium
D) None of these
Correct Answer: B
Solution :
\[{{E}_{cell}}=E_{cell}^{0}\,-\frac{0.0591}{1}{{\log }_{10}}\,\frac{{{[{{H}^{+}}]}_{anode}}}{{{[{{H}^{+}}]}_{cathode}}}\,\]\[=\frac{0.0591}{1}{{\log }_{10}}\frac{{{10}^{-2}}}{{{10}^{-3}}}\,=-ve\] (non spontaneous).
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