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Consider the given data. {:(,"Half-cel...

Consider the given data.
`{:(,"Half-cell reaction","Standard reduction"),(,,"potential"E^(@)("volts")),(1.,Cr_(2)O_(7)^(2-)+14H^(+)+6e hArr 2Cr^(3+)+7H_(2)O,1.33),(2.,Cr^(3+)+3ehArr Cr(s),-0.74),(3.,Cu^(+)+ehArrCu(s),0.51),(4.,Cu^(2+)+2ehArrCu(s),0.34):}`
The numerical value of the standard cell potential for the reaction, `Cr(s) +3Cu^(+2) rar 3Cu^(+)+Cr^(+3)(aq)` is:

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Consider the given data. {:(,"Half-cell reaction","Standard reduction"),(,,"potential"E^(@)("volts")),(1.,Cr_(2)O_(7)^(2-)+14H^(+)+6e hArr 2Cr^(3+)+7H_(2)O,1.33),(2.,Cr^(3+)+3ehArr Cr(s),-0.74),(3.,Cu^(+)+ehArrCu(s),0.51),(4.,Cu^(2+)+2ehArrCu(s),0.34):} The numerical value of the standard cell potential for the reaction, Cr(s) +3Cu^(+2)(aq) hArr 3Cu^(+)+Cr^(+3)(aq) is:

What is the standard cell potential for the reaction, 2Cr(s) + 3Sn^(2+)(aq) rightarrow 3Sn(s) + 2Cr^(3+)(aq) given the E^(@) values shown?

For reaction : Cr_(2)O_(7)^(-2)+14H^(+)rarr2Cr^(+3)+7H_(2)O , How many e^(-)s are required

For reaction : Cr_(2)O_(7)^(-2)+14H^(+)rarr2Cr^(+3)+7H_(2)O , How many e^(-)s are required

The electrode with reaction :Cr_(2)O_(7)^(2-)(aq)+14H^(o+)(aq)+6e^(-) rarr 2Cr^(3+)(aq)+7H_(2)O can be represented as

The electrode with reaction :Cr_(2)O_(7)^(2-)(aq)+14H^(o+)(aq)+6e^(-) rarr 2Cr^(3+)(aq)+7H_(2)O can be represented as

Cr_(2)O_(7)^(2-) + I^(-) rarr I_(2)+Cr^(3+) E_(cell)^(@)=0.79V,E_(Cr_(2)O_(7)^(2-))^(@)=1.33V . Then E_(I_(2))^(@)=?

A solution contaning 4.5 mM of Cr_(2)O_(7)^(2-) and 15 mM of Cr^(3+) shows a pH of 2.0 . Calculater the potential of half reaction. (Standard potential of the reaciton Cr_(2)O_(7)^(2-) rarr Cr^(3+) is 1.33V )

For Cr_2O_7^(2-) + 14 H^(+) + 6e^(-) to 2Cr^(+3) + 7H_2O, E^(@) = 1.33 V " At" [ Cr_2O_7^(2-)]=4.5 millimoles , [Cr^(+3)]=15 millimole , E is 1.067V. The pH of the solution is nearly equal to