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7. Balance the following reaction under basic conditions: Al (s) + Cr2O72- (aq) -> Al3+ (aq) + Cr3+ (aq)

8. What is the ^o of the following balanced redox reaction (use the reduction potentials provided in Table 11.1
below)?

Fe3+ (aq) + Au (s) -> Fe (s) + Au3+ (aq)

9. What is G for the reaction below – calculate it from the ^o ?

Fe (s) + Au3+ (aq) -> Fe3+ (aq) + Au (s)

7 Balance The Following Reaction Under Basic Conditions Al S Cr2O72 Aq Gt Al3 Aq Cr3 Aq 8 What Is The O Of The Following Balanced Redox Reaction Use The Reducti class=

Answer :

PSTNONSONJOKUGO

The free energy change of the reaction; Fe (s) + Au3+ (aq) -> Fe3+ (aq) + Au (s) is calculated to be -443.83KJ/mol.

For the reaction shown in question 7, we can divide it into half equations as follows;

Oxidation half equation;

6 Al (s) -------> 6Al^3+(aq) + 18e

Reduction half equation;

3Cr2O7^2-(aq) + 42H^+(aq)   + 18e -----> 6Cr^3+(aq) + 21H2O(l)

The balanced reaction equation is;

6Al(s) + 3Cr2O7^2-(aq) + 42H^+(aq)   -----> 6Al^3+(aq) + 6Cr^3+(aq) + 21H2O(l)

The E° of this reaction is obtained from;

E° anode = -0.04 V

E°cathode = +1.50 V

E° cell = +1.50 V - (-0.04 V) = 1.54 V

Given that;

ΔG° = -nFE°cell

n = 3, F = 96500, E°cell = 1.54 V

ΔG° = -(3 × 96500 × 1.54)

ΔG° = -443.83KJ/mol

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