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Calcualte the emf of the following cell ...

Calcualte the emf of the following cell at 298 K:
`2Cr(s) + 3Fe^(2+) (0.1 M ) to 2Cr^(3+) (0.01 M) + 3Fe(s)`
Given: `E_("Cr^(3+)//Cr)^(@) = -0.74 V, E_(Fe^(2+)//Fe)^(@) = -0.44 V`.

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Verified by Experts

Standard emf of the Cr - Fe cell is calculated as under:
`E_("cell")^(@) = E_("cathode")^(@) - E_("anode")^(@)`
`=-0.44 - (-0.74) = 0.30 V`
`E_("cell") = E_("cell")^(@) - 0.059/n log ([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
Substituting the values, we have,
`E_("cell") = 0.30 - 0.059/6 log ([0.01]^(2))/([0.1]^(3)) = 0.30 -(-0.059//6) = 0.3098 V`
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