IO3^-(aq) + Re(s) >>> ReO4^-(aq) + IO^-(aq)
Balance half reactions ;
IO3^-(aq) >>> IO^-(aq)
balance the oxygen by adding H2O(l) and then balance hydrogens by adding H^+(aq) in opposite side ;
IO3^-(aq) + 4H^+(aq) >>>> IO^-(aq) + 2H2O(l)
balance charge by electrons (e^-) ;
IO3^-(aq) + 4H^+(aq) + 4e^- >>>> IO^-(aq) + 2H2O(l)
Above is balanced first half reaction.
Now second half ; Re(s) >>> ReO4^-(aq)
Similarly here , balance the oxygen and then hydrogen.
Re(s) + 4H2O(l) >>>> ReO4^-(aq) + 8H^+(aq)
balance charge by e^- ;
Re(s) + 4H2O(l) >>>> ReO4^-(aq) + 8H^+(aq) + 7e^-
Above is second half balanced equation.
Now in net ionic equation there should be no electrons , so multiply first half balanced with 7 and second half balanced with 4 to balance and cancel the electrons and then add up the multiplied equations as ;
7IO3^-(aq) + 28H^+(aq) + 16H2O(l) + 4Re(s) >>>> 7IO^-(aq) + 14H2O(l) + 4ReO4^-(aq) + 32H^+(aq)
Now remove the common H2O(l) and H^+(aq) from both sides to get balanced ionic equation of acidic medium as ;
7IO3^-(aq) + 2H2O(l) + 4Re(s) >>> 7IO^-(aq) + 4ReO4^-(aq) + 4H^+(aq)
As we have to balance the equation in basic medium then add equal number of OH^-(aq) on both sides as the number of H^+(aq) in above equation so that H^+(aq) + OH^-(aq) becomes H2O(l) and then you will remove common water (H2O) to get final balanced ionic equation in basic medium (having OH^-(aq) in equation) as ;
7IO3^-(aq) + 4Re(s) + 4OH^-(aq) >>> 7IO^-(aq) + 4ReO4^-(aq) + 2H2O(l)
Above is your required equation.
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