${Cu}({s})\left|{Cu}^{2+}({aq})(0.1 {M}) \| {Ag}^{+}({aq})(0.01 {M})\right| {Ag}({s})$ કોષ માટે,કોષનો પોટેન્શિયલ ${E}_{1}=0.3095\, {~V}$

${Cu}({s})\left|{Cu}^{2+}({aq})(0.01 {M}) \| {Ag}^{+}({aq})(0.001 {M})\right| {Ag}({s})$ કોષ માટે ,કોષનો પોટેન્શિયલ $=.....\times 10^{-2} {~V}$

[ઉપયોગ : $\frac{2.303 {RT}}{{F}}=0.059$ ]

Question

${Cu}({s})\left|{Cu}^{2+}({aq})(0.1 {M}) \| {Ag}^{+}({aq})(0.01 {M})\right| {Ag}({s})$ કોષ માટે,કોષનો પોટેન્શિયલ ${E}_{1}=0.3095\, {~V}$

${Cu}({s})\left|{Cu}^{2+}({aq})(0.01 {M}) \| {Ag}^{+}({aq})(0.001 {M})\right| {Ag}({s})$ કોષ માટે ,કોષનો પોટેન્શિયલ $=.....\times 10^{-2} {~V}$

[ઉપયોગ : $\frac{2.303 {RT}}{{F}}=0.059$ ]

JEE MAIN 2021, Diffcult

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Solution

c
Cell reaction is:

${Cu}({s})+2 {Ag}^{+}({aq}) \rightarrow {Cu}^{2+}+2 {Ag}({s})$

Now, ${E}_{\text {cell }}-\frac{0.059}{2} \log \frac{\left[{cu}^{+}+\right]}{\left[{Ag}^{+}\right]^{2}}...(1)$

$\therefore {E}_{1}=0.3095={E}_{\text {cell }}^{0}-\frac{0.059}{2} \cdot \log \frac{0.01}{(0.001)^{2}} \ldots(2)$

From $(1)$ and $(2),$ ${E}_{2}=0.28\, {~V}=28 \times 10^{-2}\, {~V}$

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