Question Details: For the reaction 2 Cr2+ (aq) + Cl2 (g) == 2 Cr 3+(aq) + 2 Cl-(

Question

Question Details: For the reaction 2 Cr2+ (aq) + Cl2 (g) == 2 Cr 3+(aq) + 2 Cl-(aq),
the value of E (cell) is 1.78 volt. What is the value of E cell forCr3+ (aq) + Cl- (aq) == Cr 2+(aq) + 1/2 Cl2 (g)
helllllllllllllllp
possible answers
-1.78 .89 1.78 -.89 -3.56 volt
2. What is the standard free energy for the cell reaction ofhis galvanic cell?
3. The actual concentrations are : Co2+ (aq) = 0.0100M, Cr 3+= 0.00100M. What is the actual potential of this galvaniccell? Question Details:
the value of E (cell) is 1.78 volt. What is the value of E cell forCr3+ (aq) + Cl- (aq) == Cr 2+(aq) + 1/2 Cl2 (g)
helllllllllllllllp
possible answers
-1.78 .89 1.78 -.89 -3.56 volt
2. What is the standard free energy for the cell reaction ofhis galvanic cell?
3. The actual concentrations are : Co2+ (aq) = 0.0100M, Cr 3+= 0.00100M. What is the actual potential of this galvaniccell? Question Details: For the reaction 2 Cr2+ (aq) + Cl2 (g) == 2 Cr 3+(aq) + 2 Cl-(aq),
the value of E (cell) is 1.78 volt. What is the value of E cell forCr3+ (aq) + Cl- (aq) == Cr 2+(aq) + 1/2 Cl2 (g)
helllllllllllllllp
possible answers
-1.78 .89 1.78 -.89 -3.56 volt
2. What is the standard free energy for the cell reaction ofhis galvanic cell?
3. The actual concentrations are : Co2+ (aq) = 0.0100M, Cr 3+= 0.00100M. What is the actual potential of this galvaniccell?

Explanation / Answer

   1. Cell potential is an intensive property(does not depend on the amount of reactants and producs).         When the two halfcells are intrechanged the cell potential takes negative sign. sothe possible answer is -1.78V 2. WE know that G0 =-nFE0cell                                    = - (2mol)(96485C/mol)(1.78V)                                    = -343532.88J 3.   Ecell = E0cell-(0.0591V/n)log{[Cr^2+]2[Cl^-]2/[Cr^3+]2}                     E0cell= +0.41V and [Cr2+ (aq)] = 0.0100 M, [Cr +3]= 0.00100M , [Cl-]= ?                     n= 1             plugg in the values in the above equation to get the E cell . 3.   Ecell = E0cell-(0.0591V/n)log{[Cr^2+]2[Cl^-]2/[Cr^3+]2}                     E0cell= +0.41V and [Cr2+ (aq)] = 0.0100 M, [Cr +3]= 0.00100M , [Cl-]= ?                     n= 1             plugg in the values in the above equation to get the E cell .

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