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Exercise · Q12

Q.Balance the following ionic equation in acidic medium by the ion-electron method: Zn+NO3−→Zn2++NH4+\text{Zn} + \text{NO}_3^{-} \to \text{Zn}^{2+} + \text{NH}_4^{+}.

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Reduction half: NO3−→NH4+\text{NO}_3^{-}\to\text{NH}_4^{+}; N goes from +5+5 to −3-3, an 8-electron reduction. Balance O with 3H2O3\text{H}_2\text{O} on the right: NO3−→NH4++3H2O\text{NO}_3^{-}\to\text{NH}_4^{+}+3\text{H}_2\text{O}; this needs 1010 H on the right (4 in NH4+\text{NH}_4^+ plus 6 in 3H2O3\text{H}_2\text{O}) balanced by 10H+10\text{H}^{+} on the left: NO3−+10H+→NH4++3H2O\text{NO}_3^{-}+10\text{H}^{+}\to\text{NH}_4^{+}+3\text{H}_2\text{O}; balance charge (left −1+10=+9-1+10=+9, right +1+1) with 8e−8e^{-} added to the left: NO3−+10H++8e−→NH4++3H2O\text{NO}_3^{-}+10\text{H}^{+}+8e^{-}\to\text{NH}_4^{+}+3\text{H}_2\text{O}.\n\nOxidation half: Zn→Zn2++2e−\text{Zn}\to\text{Zn}^{2+}+2e^{-}.\n\nCombine: LCM of 8 and 2 is 8; multiply the oxidation half by 4: 4Zn→4Zn2++8e−4\text{Zn}\to4\text{Zn}^{2+}+8e^{-}. Add and cancel the 8e−8e^-: $\text{ …

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