In this case, the problem is asking for the balance of a redox reaction in acidic media, in which nickel is reduced to a metallic way and nitrogen oxidized to an ionic way.
Thus, according to the given information, it turns out possible for us to balance this equation in acidic solution by firstly setting up the half reactions:
[tex]Ni^{2+}+2e^-\rightarrow Ni^0\\\\N^{3-}H_4^++3H_2O\rightarrow N^{5+}O_3^-+8e^-+10H^+[/tex]
Next, we cross multiply each half-reaction by the other's carried electrons:
[tex]8Ni^{2+}+16e^-\rightarrow 8Ni^0\\\\2N^{3-}H_4^++6H_2O\rightarrow 2N^{5+}O_3^-+16e^-+20H^+[/tex]
Finally, we add them together to obtain:
[tex]8Ni^{2+}+2N^{3-}H_4^++6H_2O\rightarrow 8Ni^0+2N^{5+}O_3^-+20H^+[/tex]
Which can be all simplified by a factor of 2 to obtain:
[tex]4Ni^{2+}+N^{3-}H_4^++3H_2O\rightarrow 4Ni^0+N^{5+}O_3^-+10H^+\\\\4Ni^{2+}(aq)+NH_4^+(aq)+3H_2O(l)\rightarrow 4Ni(s)+NO_3^-(aq)+10H^+(aq)[/tex]
Hence, the coefficients in front of Ni and H⁺ are 4 and 10 respectively.
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