A) +1.19 V
B) +0.89 V
C) +0.18 V
D) +1.83 V
Correct Answer: B
Solution :
[b] Given \[{{E}_{S{{n}^{4+}}/S{{n}^{2+}}}}=+0.15V\] \[{{E}_{C{{r}^{3+}}/Cr}}=-0.74V\] \[E_{cell}^{{}^\circ }=E_{cathode}^{{}^\circ }-E_{anode}^{{}^\circ }\] \[=0.15-\left( -0.74 \right)=+0.89V\]You need to login to perform this action.
You will be redirected in
3 sec