fe(s)+niso₄(aq) → feso₄(aq)+ni(s)\nwrite a balanced equation for the half - reaction that happens at the…

Explanation:

Step1: Identify reduction and oxidation half - reactions

In a galvanic cell, reduction occurs at the cathode and oxidation at the anode. For the reaction $Fe(s)+NiSO_4(aq)\to FeSO_4(aq) + Ni(s)$, nickel is reduced and iron is oxidized.

Step2: Write cathode half - reaction

The reduction of nickel ions occurs at the cathode. The half - reaction is $Ni^{2 + }(aq)+2e^-\to Ni(s)$.

Step3: Write anode half - reaction

The oxidation of iron occurs at the anode. The half - reaction is $Fe(s)\to Fe^{2+}(aq)+2e^-$.

Step4: Determine electrode materials

The anode is made of the metal that is oxidized. So, $E1$ (anode) is made of iron ($Fe$). The cathode is made of the metal that is reduced. So, $E2$ (cathode) is made of nickel ($Ni$).

Step5: Identify chemical species in solutions

In solution $S1$ (where oxidation occurs), the chemical species are $Fe^{2+}$ ions from the oxidation of iron and the sulfate ions ($SO_4^{2 - }$) from $FeSO_4$. In solution $S2$ (where reduction occurs), the chemical species are $Ni^{2+}$ ions from $NiSO_4$ and $SO_4^{2 - }$ ions.

Answer:

1. Cathode half - reaction: $Ni^{2 + }(aq)+2e^-\to Ni(s)$
2. Anode half - reaction: $Fe(s)\to Fe^{2+}(aq)+2e^-$
3. $E1$: Iron ($Fe$)
4. $E2$: Nickel ($Ni$)
5. $S1$: $Fe^{2+},SO_4^{2 - }$
6. $S2$: $Ni^{2+},SO_4^{2 - }$