The standard EMF of a Daniel cell at 298 K is . When the concentration of ZnSO is 1.0 M and that of — Electrochemistry Chemistry Question
Question
The standard EMF of a Daniel cell at 298 K is $E_1$. When the concentration of ZnSO$_4$ is 1.0 M and that of CuSO$_4$ is 0.01 M, the EMF becomes $E_2$ at 298 K. The correct relationship between $E_1$ and $E_2$ is
Answer: C
💡 Solution & Explanation
$E_2 = E_1 - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]} = E_1 - 0.0295 \log \frac{1.0}{0.01} = E_1 - 0.0295 \log 100 = E_1 - 0.0591$. Since a positive value is subtracted from $E_1$, $E_1$ must be greater than $E_2$. Therefore, correct answer is C.
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