Given the equilibrium constant: K_ C of the reaction: Cu(s)+2 Ag^ + (aq) arrow Cu^ 2+ (aq)+2 Ag(s) is 10 × 10^ 15 calculate the E_ cell ^ 0 of this reaction at 298 ~K[2.303 RT F . at .298 ~K=0.059 ~V]

Correct answer: C. 0.4736 ~V E_ cell ^ 0 = 2.303 RT nF K_ C or E_ cell ^ 0 = 0.059 ~V n K_ C = 0.059 ~V 2 10^ 16 =0.4736 ~V

JEE Main 201911 JanuaryEvening Shift

JEE Main 2019 · 11 January · Evening Shift · Chemistry · Electrochemistry

Given the equilibrium constant: K_ C of the reaction: Cu(s)+2 Ag^ +...

Given the equilibrium constant:

\mathrm{K}_{\mathrm{C}}

of the reaction:

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

10 \times 10^{15}

calculate the

E_{\text{cell }}^{0}

of this reaction at

298 \mathrm{~K}\left[2.303 \frac{\mathrm{RT}}{\mathrm{F}}\right.

at \left.298 \mathrm{~K}=0.059 \mathrm{~V}\right]

C. $0.4736 \mathrm{~V}$

E_{\text{cell }}^{0}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{C}}

\mathrm{E}_{\text{cell }}^{0}=\frac{0.059 \mathrm{~V}}{\mathrm{n}} \log \mathrm{K}_{\mathrm{C}}=\frac{0.059 \mathrm{~V}}{2} \log

10^{16}=0.4736

\mathrm{~V}