Need Help on an Electrochem Q
Use half-equations from the table to deduce an equation for the reduction of VO2^( ) to VO^(2 ) in aqueous solution by Iron. [Fe(H2O)6]^2 (aq) 2e- --> Fe (s) 6H2O (l) E' = -0.44VH (aq) e- --> 1/2H2 (g) E' = 0.00V[Fe(H2O)6]^3 (aq) e- --> [Fe(H2O)6]^2 (aq) E' = 0.77VVO2^( ) (aq) 2H (aq) e- --> VO^(2 ) (aq) H2O (l) E' = 1.00V The answer I got was:2VO2^( ) 4H Fe 4H2O ---> 2VO^(2 ) [Fe(H2O)6]^2 But the correct answer on the mark scheme is:3 VO2^( ) 6 H Fe 3 H2O → 3 VO^(2 ) [Fe(H2O)6]^3 Can someone please explain to me how to arrive at the correct answer? It seems I have combined the wrong half-equations together but I'm not sure why Fe3 is used and not Fe2 and why this would mean that the VO2 equation is multiplied by 3 instead of 2. Thankyou in advance 
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