A) 0.38V
B) 0.53V
C) 0.19V
D) 0.49V
Correct Answer: B
Solution :
\[C{{u}^{++}}+{{e}^{-}}\xrightarrow[{}]{{}}C{{u}^{+}}\] \[E_{1}^{0}=0.15V;\Delta G_{1}^{0}=-{{n}_{1}}E_{1}^{0}F\] \[C{{u}^{2+}}+2e\xrightarrow[{}]{{}}Cu\] \[E_{2}^{0}=0.34V;\Delta G_{2}^{0}=-{{n}_{2}}E_{2}^{0}F\] On subracting eq.(i) from eq. (ii) we get \[C{{u}^{+}}+{{e}^{-}}\xrightarrow[{}]{{}}Cu;\Delta {{G}^{0}}=\Delta G_{2}^{0}-\Delta G_{1}^{0}\] \[-n{{E}^{0}}F=-({{n}_{2}}{{E}^{0}}F-{{n}_{1}}F_{1}^{2}F)\] \[{{E}^{0}}=\frac{{{n}_{2}}E_{2}^{0}F-{{n}_{1}}E_{1}^{0}F}{nF}\] \[=\frac{2\times 0.34-0.15}{1}\] \[=0.53V\]
You need to login to perform this action.
You will be redirected in
3 sec
