Balancing redox reactions in basic solutions example

Let’s balance the following redox reaction in basic conditions.

\inline \fn_jvn \large {Ag}_{(s)}+{Zn^{2+}}_{(aq)}\rightleftharpoons {Ag_{2}O}_{(aq)}+{Zn}_{(s)}

Solution

Step 1: Separate the half-reactions.

\inline \fn_jvn \large {Ag}_{(s)}\rightarrow {Ag_2O }_{(aq)}

$\inline \fn_jvn \large {Zn^{2+}}_{(aq)} \rightarrow {Zn}_{(s)}$

Step 2: Balance elements other than O and H.

\inline \fn_jvn \large {2Ag}_{(s)} \rightarrow {Ag_2O}_{(aq)}

$\inline \fn_jvn \large {Zn^{2+}}_{(aq)} \rightarrow {Zn }_{(s)}$

Step 3: Add H₂O to balance oxygen.

\inline \fn_jvn \large {H_2O}_{(l)} + {2Ag}_{(s) }\rightarrow {Ag_{2}O}_{(aq)}

$\inline \fn_jvn \large {Zn^{2+}}_{(aq)} \rightarrow {Zn}_{(s)}$

Step 4: Balance hydrogen with protons.

\inline \fn_jvn \large {H_{2}O }_{(l)} + {2Ag}_{(s)} \rightarrow {Ag_{2}O}_{(aq)} + {2H^{+}}_{(aq)}

$\inline \fn_jvn \large {Zn^{2+} }_{(aq)} \rightarrow {Zn }_{(s)}$

Step 5: Balance the charge with e^–.

Step 6: Scale the reactions so that they have an equal amount of electrons. In this case, it is already done.

$\inline \fn_jvn \large {H_{2}O }_{(l)} + {2Ag }_{(s)} \rightarrow {Ag_{2}O}_{(aq)} + {2H^{+}}_{(aq)} + 2e^-$

$\inline \fn_jvn \large {Zn^{2+}}_{(aq)} + 2e^- \rightarrow {Zn}_{(s)}$

Step 7: Add the reactions and cancel the electrons.

\inline \fn_jvn \large {H_{2}O}_{(l)} + {2Ag}_{(s)}+ {Zn^{2+}}_{(aq)} \rightarrow {Zn}_{(s)} + {Ag_{2}O}_{(aq)} + {2H^+}_{(aq)}

Step 8: Add OH^– to balance H⁺. There are 2 net protons in this equation, so add 2 OH^– ions to each side.

$\inline \dpi{120} \fn_jvn {H_2O}_{(l) }+ {2Ag}_{(s)} + {Zn^{2+}}_{(aq)} + {2OH^-}_{(aq)} \rightarrow {Zn}_{(s)} + {Ag_2O}_{(aq)} + {2H^+}_{(aq)} + {2OH^-}_{(aq)}$

Step 9: Combine OH^– ions and H⁺ ions that are present on the same side to form water.

\inline \dpi{120} \fn_jvn {H_2O}_{(l)} + {2Ag}_{(s)} + {Zn^{2+}}_{(aq)} + {2OH^{-}}_{(aq)} \rightarrow {Zn}_{(s)} + {Ag_{2}O}_{(aq)} + {2H_{2}O}_{(l)}

Step 10: After cancelling common terms, we have.

$\inline \dpi{120} \fn_jvn \large {2Ag}_{(s)} + {Zn^{2+}}_{(aq)} + {2OH^- }_{(aq)} \rightarrow {Zn}_{(s)} + {Ag_{2}O}_{(aq)} + {H_{2}O}_{(l)}$

Neutral conditions example

Balance the following reaction

\inline \fn_jvn \large {Cu^+}_{(aq)} + {Fe}_{(s)} \rightarrow {Fe^{3+} }_{(aq)} + {Cu }_{(s)}

Solution

Step 1: Separate the half-reactions. By searching for the reduction potential, one can find two separate reactions:

Cu + (aq) + e - \to Cu (s)

{Cu}^{+} (aq) + e^{-} \to Cu (s)

{Fe}^{3 +} (aq) + 3 e^{-} \to Fe (s)

The copper reaction has a higher potential and thus is being reduced. Iron is being oxidized so the half-reaction should be flipped. This yields:

Cu + (aq) + e - \to Cu (s)

{Cu}^{+} (aq) + e^{-} \to Cu (s)

Fe (s) \to {Fe}^{3 +} (aq) + 3 e^{-}

Step 2: Balance the electrons in the equations. In this case, the electrons are simply balanced by multiplying the entire $C u^{+} (a q) + e^{-} \to C u (s)$ half-reaction by 3 and leaving the other half reaction as it is. This gives: