\[C{{u}^{+}}+{{e}^{-}}\xrightarrow{{}}Cu,\]\[E{}^\circ ={{x}_{1}}\,\,\text{volt}\,\text{;}\] |
\[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu,\]\[E{}^\circ ={{x}_{2}}\,\,\text{volt,}\] then for |
\[C{{u}^{2+}}+{{e}^{-}}\xrightarrow{{}}C{{u}^{+}},\]\[E{}^\circ \,\,(\text{volt)}\] will be - |
A) \[{{x}_{1}}-2{{x}_{2}}\]
B) \[{{x}_{1}}+2{{x}_{2}}\]
C) \[{{x}_{1}}-{{x}_{2}}\]
D) \[2{{x}_{2}}-{{x}_{1}}\]
Correct Answer: D
Solution :
\[C{{u}^{+}}+{{e}^{-}}\xrightarrow{{}}Cu,\,\,E'={{x}_{1}}\,\,\text{Volt}\] |
\[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu,\,\,={{x}_{2}}\,\,\text{Volt}\] |
\[Cu\xrightarrow{{}}C{{u}^{+}}+{{e}^{-}}-{{x}_{1}}\,\,\text{Volt}\] |
\[C{{u}^{2+}}+{{e}^{-}}\xrightarrow{{}}C{{u}^{+}}\] |
\[-\,2\times {{x}_{2}}\times f+1\times {{x}_{1}}\times f=-1\times {{E}^{0}}\times f\] |
\[{{E}^{0}}=2{{x}_{2}}-{{x}_{1}}\] |
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