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The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is $Fe^{2+}_{(aq)}+2e^-\rightarrow Fe(s),\ E^\circ=-0.44$ V; $Cr_2O_{7\ (aq)}^{2-}+14H^++6e^-\rightarrow2Cr^{3+}+7H_2O,\ E^\circ=+1.33$ V
A
+1.77 V
B
+2.65 V
C
+0.01 V
D
+0.89 V
Detailed Solution
$E^\circ_{cell}=E^\circ_C-E^\circ_A=(1.33)-(-0.44)=+1.77$ V.