Question

An electrochemical cell has two half cell reactions as, $A^{2+}+2 e^{-} \rightarrow A ; E_{A^{2+} / A}^{o}=0.34\, V$
$X \rightarrow X^{2+}+2 e^{-} ; E_{X^{2+} / X}^{o}=-2.37\, V$
The cell voltage will be

• A. 2.71 V
• B. 2.03 V
• C. -2.71 V
• D. -2.03 V

Answer 1

Answer: (A)

Given, $E_{A^{2+} / A}^{o}=0.34\, V,\, E_{x^{2+} / x}^{o}=-2.37\, V$
Cell voltage $=E_{\text {cathode }}^{o}-E_{\text {Anode }}^{o}$
$=0.34-(-2.37)=2.71\, V$