`Cr_(2)O_(7)^(2-)+I^(-)toI_(2)+Cr^(3+)` `E_(cell)^(@)=0.79V,E_(Cr_(2)O_(7)^(2-))^(@)=1.33V,E_(I_(2))^(@)=?`A. `0.54V`B. `-0.054V`C. `+0.18V`D. `-0.18V`

`Cr_(2)O_(7)^(2-)+I^(-)toI_(2)+Cr^(3+)` `E_(cell)^(@)=0.79V,E_(Cr_(2)O

1.	`Cr_(2)O_(7)^(2-)+I^(-)toI_(2)+Cr^(3+)` `E_(cell)^(@)=0.79V,E_(Cr_(2)O_(7)^(2-))^(@)=1.33V,E_(I_(2))^(@)=?`A. `0.54V`B. `-0.054V`C. `+0.18V`D. `-0.18V`
Answer» Correct Answer - D In the given raction, `I^(-)` has been oxidized to `I_(2)` and `Cr_(2)O_(7)^(2-)` ions have been reduced to `Cr^(3+)` `thereforeE_(cell)^(@)=E_(Cr_(2)O_(7)^(2-))^(@)-E_(I_(2))^(@)` i.e., `0.79=1.33-E_(I_(2))^(@)` or `E_(I_(2))^(@)=0.54V`

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`Cr_(2)O_(7)^(2-)+I^(-)toI_(2)+Cr^(3+)` `E_(cell)^(@)=0.79V,E_(Cr_(2)O_(7)^(2-))^(@)=1.33V,E_(I_(2))^(@)=?`A. `0.54V`B. `-0.054V`C. `+0.18V`D. `-0.18V`

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