| \[Cr|C{{r}^{3+}}(1.0M)||C{{o}^{2+}}(1.0M)|Co\] |
| \[(E{}^\circ for\text{ }C{{r}^{3+}}/Cr=-0.74\] volt and \[E{}^\circ \] for \[C{{o}^{2+}}/Co=-0.28\] volt) |
A) \[-0.46\]volt
B) \[-1.02\,~volt\]
C) \[+0.46\] volt
D) 1.66 volt
Correct Answer: C
Solution :
[c] \[E_{C{{r}^{3+}}/Cr}^{{}^\circ }=-0.74V,\,E_{C{{o}^{2+}}/Co}^{{}^\circ }=-0.28V\] The given cell reaction is \[Cr|C{{r}^{3+}}(1.0M)||C{{o}^{2+}}(1.0M)|Co\] \[\therefore \text{ }Cr\] is anode and Co is cathode \[E_{cell}^{{}^\circ }=E_{C}^{{}^\circ }-E_{A}^{{}^\circ }=-0.28-(-0.74)\] \[=-0.28+0.74=+0.46V\]
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