| Emf of the cell |
| \[Ni|N{{i}^{2+}}(0.1\,M)||A{{u}^{3+}}\,(1.0\,M)|\,Au\] will be |
| \[(E_{Ni/N{{i}^{2+}}}^{{}^\circ }=0.25,\,E_{Au/A{{u}^{3+}}}^{{}^\circ }=1.5\,V)\] |
A) 1.75 V
B) \[+\]1.7795 V
C) \[+\]0.7795 V
D) \[-\]1.7795 V
Correct Answer: B
Solution :
| [b] Cell reaction: |
| \[3Ni+2A{{u}^{+3}}\xrightarrow{{}}3N{{i}^{+2}}+2Au\] |
| \[{{E}_{cell}}=E_{cell}^{{}^\circ }-\frac{0.0591}{6}\log \,\frac{{{[N{{i}^{+2}}]}^{3}}}{{{[A{{u}^{+3}}]}^{2}}}\] |
| \[=(0.25+1.5)-\frac{0.0591}{6}\log \frac{{{(0.1)}^{3}}}{{{(1)}^{2}}}\] |
| \[=1.75-\frac{0.0591}{2}\log \,(.1)\,=\,1.75+0.295=+1.7795\,V\] |
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