An electrochemical cell has two half cell reactions as, A2++2 e- arrow A ; EA2+ / Ao=0.34 V X arrow X2++2 e- ; EX2+ / Xo=-2.37 V The cell voltage will be

Question

An electrochemical cell has two half cell reactions as, A2++2 e- arrow A ; EA2+ / Ao=0.34 V X arrow X2++2 e- ; EX2+ / Xo=-2.37 V The cell voltage will be (A) 2.71 V (B) 2.03 V (C) -2.71 V (D) -2.03 V

Tardigrade Admin · Accepted Answer

Given, EA2+ / Ao=0.34 V, Ex2+ / xo=-2.37 V Cell voltage =E text cathode o-E text Anode o =0.34-(-2.37)=2.71 V

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

2.71 V

2.03 V

-2.71 V

-2.03 V

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

Q. An electrochemical cell has two half cell reactions as, A2++2e−→A;EA2+/Ao​=0.34V X→X2++2e−;EX2+/Xo​=−2.37V The cell voltage will be