write balanced half - reactions for the following redox reaction: 6cu^{2 + }(aq)+2cr^{3 + }(aq)+7h_{2}o(l)→6c…

Explanation:

Step1: Identify reduction half - reaction

Reduction is the gain of electrons. Copper goes from $Cu^{2 +}$ to $Cu^{+}$. $Cu^{2+}(aq)+e^{-}\to Cu^{+}(aq)$ Multiply by 6 to match the number of copper ions in the overall reaction: $6Cu^{2+}(aq)+6e^{-}\to 6Cu^{+}(aq)$

Step2: Identify oxidation half - reaction

Oxidation is the loss of electrons. Chromium goes from $Cr^{3 +}$ to $Cr_{2}O_{7}^{2 -}$. First, balance the chromium atoms: $2Cr^{3+}(aq)\to Cr_{2}O_{7}^{2 -}(aq)$ Then balance oxygen by adding water: $2Cr^{3+}(aq)+7H_{2}O(l)\to Cr_{2}O_{7}^{2 -}(aq)$ Balance hydrogen by adding $H^{+}$: $2Cr^{3+}(aq)+7H_{2}O(l)\to Cr_{2}O_{7}^{2 -}(aq)+14H^{+}(aq)$ Finally, balance the charge by adding electrons: $2Cr^{3+}(aq)+7H_{2}O(l)\to Cr_{2}O_{7}^{2 -}(aq)+14H^{+}(aq)+6e^{-}$

Answer:

reduction: $6Cu^{2+}(aq)+6e^{-}\to 6Cu^{+}(aq)$ oxidation: $2Cr^{3+}(aq)+7H_{2}O(l)\to Cr_{2}O_{7}^{2 -}(aq)+14H^{+}(aq)+6e^{-}$