Solve this: Q13 An electrochemical cell has two half cell reactions as, - Chemistry - Redox Reactions

Question

Solve this:

Q13 An electrochemical cell has two half cell reactions as,

A 2 +   +   2 e -   →   A   ;  
E °   = 0 34   V X   +   X 2 +  
→   2 e -   ;   E °   = + 2 37  
V

The cell voltage will be

(1)2 71 V (2) 2 03 V

(3) – 2 71 V (4) – 2 03 V

Decoder_12 · Accepted Answer

Dear student,

The oxidation reaction given above seems to be incorrect, so i am
assuming X  → X2+ + 2e- in place of
X + X2+ → 2e-
If you are still having any doubts about the topic then you can
kindly repost your question

Now, for calculating the cell potential/voltage we can use the
relation :

Ecello = Ereductiono + Eoxidationo
------(1)
So, in this case we are provided with reduction potential of
both the half reactions,
i e
EA2+/Ao (reduction potential of A2+) = +0
34VEX2+/Xo (reduction potential of X) = +2
37V

We know that, oxidation is reverse of reduction, so;

Eoxidationo=-EreductionoEX/X2+o=-EX2+/XoEX/X2+o=-2 37 V
Now, using equation (1);

Ecello= EA2+/Ao + EX/X2+oEcell o = +0
34V + (-2 37V)Ecello = -2 03V

Therefore, option (4) is correct

Hope this information
will clear your doubts about the topic

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have any more doubts just ask here on the forum and our experts
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Regards

Solve this: Q13. An electrochemical cell has two half cell reactions as, A 2 + + 2 e - → A ; E ° = 0 . 34 V X + X 2 + → 2 e - ; E ° = + 2 . 37 V The cell voltage will be (1)2.71 V (2) 2.03 V (3) – 2.71 V (4) – 2.03 V

Solve this:

Q13. An electrochemical cell has two half cell reactions as,

$A^{2 +} + 2 e^{-} \to A; E ° = 0.34 V X + X^{2 +} \to 2 e^{-}; E ° = + 2.37 V$

The cell voltage will be

(1)2.71 V (2) 2.03 V

(3) – 2.71 V (4) – 2.03 V