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Introduction

An electroche...

Question

An electrochemical cell has two half cell reactions as,
$Y^{3 +} + 3 e^{-} \to Y : E = 0.22 V$
$Z \to Z^{3 +} + 3 e^{-} : E = 2.27 V$
Calculate the $E_{c e l l}$ .

A

2.99 V .

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B

2.49 V .

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C

2.05 V .

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D

- 2.49 V .

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Solution

The correct option is B $2.49 V .$
Given reactions are:

$Y^{3 +} + 3 e^{-} \to Y : E^{o} = 0.22 V$
$Z \to Z^{3 +} + 3 e^{-} : E^{o} = 2.27 V$

We know that,
$E_{c e l l} = E_{c a t h o d e} - E_{a n o d e}$

Here,
$E_{c a t h o d e} = 0.22 V$
$E_{a n o d e} = - 2.27 V$ , because initially oxidation potential is given, so to get reduction potential sign is changed.

So, putting the values,

$E_{c e l l} = 0.22 - (- 2.27) = 2.49 V .$

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Similar questions

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

Y^{3 +} + 3 e^{-} \to Y : E = 0.22 V

Z \to Z^{3 +} + 3 e^{-} : E = 2.27 V

E_{c e l l}

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{A g}^{+} + e^{-} ⟶ A g; E_{{A g}^{+} | A g}^{o} = - 0.3995 V

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