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What is the standard free energy change for the following redox reaction at 298K?Cr2O7(2-) + 14H(+) + 6e(-) → 2Cr(3+) + 7H2O(l)Given: E°(Cr2O7(2-)/Cr(3+)) = 1.33 V and E°(H2O/O2) = -0.82 V.

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To calculate the standard free energy change  G  for the given redox reaction, we can use the Nernst equation:G = -nFEwhere n is the number of electrons transferred in the reaction, F is the Faraday constant  96,485 C/mol , and E is the standard cell potential.First, we need to determine the number of electrons transferred in the reaction. In this case, the reaction involves the transfer of 6 electrons.Next, we need to determine the standard cell potential  E  for the reaction. Since we are given the standard reduction potentials for both half-reactions, we can calculate the overall cell potential by adding them together:E Cr2O7 2- /Cr 3+   = 1.33 VE H2O/O2  = -0.82 VE cell  = E Cr2O7 2- /Cr 3+   + E H2O/O2  = 1.33 V +  -0.82 V  = 0.51 VNow we can calculate the standard free energy change  G :G = -nFE = -6 mol * 96,485 C/mol * 0.51 V = -295,861.9 J/molSince we typically express free energy changes in kJ/mol, we can convert this value:G = -295,861.9 J/mol *  1 kJ/1000 J  = -295.86 kJ/molSo, the standard free energy change for the given redox reaction at 298 K is -295.86 kJ/mol.

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