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For the cell:
\mathrm{Zn}(\mathrm{s})\left|\mathrm{\: Zn}^{2+}(a \mathrm{\: M}) \| \mathrm{Ag}^{+}(b \mathrm{\: M})\right| \mathrm{Ag}(\mathrm{s})
Write Nernst equation to show how Ecell vary with concentration of Zn2+ and Ag+ ions. Given E- (Zn2+| Zn) = -0.76 V,  E-(Ag+| Ag) = 0.80 V.

Option: 1

\mathrm {E_{cell}}=E_{\text {cell }}^{-}-\frac{0.059}{2} \log \left[Z n^{2+}\right]+\frac{0.059}{2} \log \left[A g^{+}\right]^{2}


Option: 2

\mathrm {E_{cell}}=E_{\text {cell }}^{-}-\frac{0.059}{2} \log \left[Z n^{2+}\right]+\frac{0.059}{2} \log \left[A g^{+}\right]


Option: 3

\mathrm {E_{cell}}=E_{\text {cell }}^{-}+\frac{0.059}{2} \log \left[Z n^{2+}\right]-\frac{0.059}{2} \log \left[A g^{+}\right]


Option: 4

None of above


Answers (1)

best_answer

Anode reaction:
                          \mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{Zn}^{2+}(a \mathrm{\: M})+2\mathrm e^{-}
Cathode reaction:
                          2 \mathrm{\: Ag}^{+}(b \mathrm{\: M})+2 \mathrm e^{-} \longrightarrow 2 \mathrm{Ag}(\mathrm{s})
Cell reaction:
     \mathrm{Z n(s)}+2 \mathrm{A g^{+}(b \: M)} \longrightarrow \mathrm{Zn}^{2+}(\mathrm{a\: M})+2\: \mathrm{Ag}(\mathrm{s})
\\E^{-}_{\text {cell }}=\left(E^{-} \text { reduction }\right)_{c}-\left(E^{-} \text { reduction }\right)_{\mathbf{a}}\\\\E^{-}_{\text {cell }} =0.80-(-0.76)=1.56 \mathrm{\: V}
Using the Nernst equation,
\mathrm E_{\text {cell }}=\mathrm E^-_{\text {eell }}-\frac{0.059}{\mathrm n_{\text {cell }}} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}
          \mathrm E_{\text {cell }}=\mathrm E^{-}_{\text {cell }}-\frac{0.059}{2} \log \left[\mathrm{Zn}^{2+}\right]+\frac{0.059}{2} \log \left[\mathrm{Ag}^{+}\right]^{2}

\begin{array}{l}{\text { Therefore, } \mathrm E_{\text {cell}} \textup{ wil decrease when }\left[\mathrm{Zn}^{2+}\right] \text { increases }} \\ {\left[\mathrm{Ag}^{+}\right] \text { decreases. }}\end{array}

Therefore, option(1) is correct

Posted by

Kshitij

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