electrochemistry

Question: Which of the following expressions represent the emf of the above cell at ${{25}^{o}}C$

1) $E=\frac{0.0592}{2}\log \frac{{{({{a}_{P{{b}^{2+}}}})}_{2}}}{{{({{a}_{P{{b}^{2+}}}})}_{1}}}$

2) $E=\frac{0.0592}{2}\log \frac{{{({{a}_{P{{b}^{2+}}}})}_{1}}}{{{({{a}_{P{{b}^{2+}}}})}_{2}}}$

3) $E=\frac{0.0592}{2}\log \frac{{{[{{K}_{sp}}(Pb{{I}_{2}})]}^{1/3}}}{{{[{{K}_{sp}}(PbS{{O}_{4}})]}^{1/2}}}$

4) $E=\frac{0.0592}{2}\log \frac{{{K}_{sp}}(Pb{{I}_{2}})}{{{K}_{sp}}(PbS{{O}_{4}})}$.

Solution:

Explanation: No Explanation

Electrode potential Ecell Nernst equation and ECS

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