For the reaction, 2 Cr^2+ (aq) + CH_2(g) 2 Cr^3+(aq) + 2 Cl^- (aq), the value of
ID: 1021901 • Letter: F
Question
For the reaction, 2 Cr^2+ (aq) + CH_2(g) 2 Cr^3+(aq) + 2 Cl^- (aq), the value of E_cell degree is 1.78 volt. What is the value of for the reaction Cr^3+(aq) + C^- (aq) rightarrow Cr^2+(aq) + 1/2 Cl_2(g)? Using the following standard reduction potentials Zn^2+ (aq) + 2e rightarrow Zn(s) E degree = - 0.76 V Ni^2+(aq) + 2e rightarrow Ni(s) E degree = - 0.25 V Calculate the standard cell potential for the cell whose reaction is Ni^2+ (aq) + Zn(s) rightarrow Zn^2+ (aq) + Ni(s) A galvanic cell is composed of these two half-cells, with the standard reduction potentials shown: Co^2+ (aq) + 2 e^- Co(s) -0 28 volt Cd^2+ (aq) + 2e^- Cd(s) -0 40 volt What is the standard free energy change (Delta G degree) for the cell reaction of this galvanic cell?Explanation / Answer
1. The value of E°cell will be same as before (1.78V) because by reversing the reaction and multiplying by some numbers will not change anything in standard cell potential KEEP IN MIND in STANDARD CELL POTENTIAL.
2. Standard cell potential E°cell = E°cathode E°anode
In this question E°cathode (reduction) = -0.25V and E°anode (oxidation) = - (-0.76) = +0.76 V (we had to multiply this potential by negative sign because in question that given potential is reduction potential and we need oxidation potential)
Put the values in the formula E°cell = (-0.25) - (+0.76) = -1.01 V (Answer)
3. We can calculate the standard free energy change by this formula G°cell = nFE°cell
where n is number of moles of electrons per mole of products = 2
F is the Faraday constant, ~96485 C/mol and E°cell = E°cathode E°anode = (-0.28) - [-(-0.40)] = -0.68 V
Put the all values in the formula we get G°cell = - 2*96485*(-0.68 ) = 131219.6 Joule (Answer)
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