What is the potential for the cell `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe` `E^(@)Cr^(3+)// Cr=-0.74V`, `E^(@)Fe^(2+)//Fe=-0.44V`A. `+0.2606V`B. `+0.5212V`C. `+0.1303V`D. `-0.2606V`.

What is the potential for the cell `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|F

1.	What is the potential for the cell `Cr\|Cr^(3+)(0.1M)\|\|Fe^(2+)(0.01M)\|Fe` `E^(@)Cr^(3+)// Cr=-0.74V`, `E^(@)Fe^(2+)//Fe=-0.44V`A. `+0.2606V`B. `+0.5212V`C. `+0.1303V`D. `-0.2606V`.
Answer» Correct Answer - A Anode half reaction : `CrrarrCu^(3+)+3e^(-)` Cathode half reaction : `Fe^(2+)+ 2Cr^(3+)+6e^(-)` or `Fe^(2)+6e^(-)rarr3Fe` cell reaction `2Cr+3Fe^(2+)rarr2Cr^(2+)+3Fe,n=6` Now `E_("cell")=E_("cell")^(@)-(0.05916)/(6)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3))` `=[-0.44-(-0.74)]-(0.05916)/(6)"log"((0.1)^(2))/((0.01)^(3))` `=0.3-0.0394=+0.2606V`

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What is the potential for the cell `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe` `E^(@)Cr^(3+)// Cr=-0.74V`, `E^(@)Fe^(2+)//Fe=-0.44V`A. `+0.2606V`B. `+0.5212V`C. `+0.1303V`D. `-0.2606V`.

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