
How do you balance redox equations in acidic solutions?
Answer
548.4k+ views
Hint: The first step in balancing any redox reaction is deciding if it is even an oxidation-decrease reaction, which requires that species displays changing oxidation states during the reaction. To keep up charge neutrality in the example, the redox reaction will involve both a reduction segment and oxidation segments and is regularly isolated into two hypothetical half-reactions to help in understanding the reaction. This requires recognizing which component is oxidized and which element is reduced.
Complete step by step answer:
This is understand by example:
\[C{r_2}{O_7}^{2 - } + N{O_2}^ - \to C{r^{3 + }} + N{O_3}^ - \]
Step1:Two half-reactions are:
\[C{r_2}{O_7}^{2 - } \to C{r^{3 + }}\]
\[N{O_2}^ - \to N{O_3}^ - \]
Step2: Balance all atoms other than \[H\]and \[O\]
\[C{r_2}{O_7}^{2 - } \to 2C{r^{3 + }}\]
\[N{O_2}^ - \to N{O_3}^ - \]
Step3: Balance \[O\]
\[C{r_2}{O_7}^{2 - } \to C{r^{3 + }} + {H_2}O\]
\[N{O_2}^ - + 1{H_2}O \to N{O_3}^ - \]
Step4: Balance \[H\]
\[C{r_2}{O_7}^{2 - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O\]
\[N{O_2}^ - + 1{H_2}O \to N{O_3}^ - + 2{H^ + }\]
Step5: Balance Charge
\[C{r_2}{O_7}^{2 - } + 14{H^ + } + 6{e^ - } \to 2C{r^{3 + }} + 7{H_2}O\]
\[N{O_2}^ - + 1{H_2}O \to N{O_3}^ - + 2{H^ + } + 2{e^ - }\]
Step6: Equalize electron transferred
\[1 \times \left[ {C{r_2}{O_7}^{2 - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O} \right]\]
\[3 \times \left[ {N{O_2}^ - + 1{H_2}O \to N{O_3}^ - + 2{H^ + }} \right]\]
Step7: Add the two half reaction
\[C{r_2}{O_7}^{2 - } + N{O_2} + 8{H^ + } \to C{r^{3 + }} + N{O_3}^ - + 4{H_2}O\]
Step8: check mass balance
On the left: \[2Cr;13O;3N;8H\]
On the right: \[2Cr;3N;13O;8H\]
Step9: Check charge balance
On the left: \[{2^ - } + {3^ - } + {8^ + } = {3^ + }\]
On the right: \[{6^ + } + {3^ - } = {3^ + }\]
The balanced equation:
\[C{r_2}{O_7}^{2 - } + N{O_2} + 8{H^ + } \to C{r^{3 + }} + N{O_3}^ - + 4{H_2}O\]
Note: Adjusting redox reactions initially requires splitting the condition into the two half-reactions of reduction and oxidation.
All atoms aside from oxygen and hydrogen ought to be balanced first.
In acidic conditions, the oxygen atoms ought to be balanced with water, while hydrogen atoms ought to be balanced with \[{H^ + }\].
In essential conditions, the oxygen atoms ought to be balanced with \[O{H^ - }\], while the hydrogen atoms ought to be balanced with water.
The completely balanced reaction ought to have both half-reactions added back together
Complete step by step answer:
This is understand by example:
\[C{r_2}{O_7}^{2 - } + N{O_2}^ - \to C{r^{3 + }} + N{O_3}^ - \]
Step1:Two half-reactions are:
\[C{r_2}{O_7}^{2 - } \to C{r^{3 + }}\]
\[N{O_2}^ - \to N{O_3}^ - \]
Step2: Balance all atoms other than \[H\]and \[O\]
\[C{r_2}{O_7}^{2 - } \to 2C{r^{3 + }}\]
\[N{O_2}^ - \to N{O_3}^ - \]
Step3: Balance \[O\]
\[C{r_2}{O_7}^{2 - } \to C{r^{3 + }} + {H_2}O\]
\[N{O_2}^ - + 1{H_2}O \to N{O_3}^ - \]
Step4: Balance \[H\]
\[C{r_2}{O_7}^{2 - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O\]
\[N{O_2}^ - + 1{H_2}O \to N{O_3}^ - + 2{H^ + }\]
Step5: Balance Charge
\[C{r_2}{O_7}^{2 - } + 14{H^ + } + 6{e^ - } \to 2C{r^{3 + }} + 7{H_2}O\]
\[N{O_2}^ - + 1{H_2}O \to N{O_3}^ - + 2{H^ + } + 2{e^ - }\]
Step6: Equalize electron transferred
\[1 \times \left[ {C{r_2}{O_7}^{2 - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O} \right]\]
\[3 \times \left[ {N{O_2}^ - + 1{H_2}O \to N{O_3}^ - + 2{H^ + }} \right]\]
Step7: Add the two half reaction
\[C{r_2}{O_7}^{2 - } + N{O_2} + 8{H^ + } \to C{r^{3 + }} + N{O_3}^ - + 4{H_2}O\]
Step8: check mass balance
On the left: \[2Cr;13O;3N;8H\]
On the right: \[2Cr;3N;13O;8H\]
Step9: Check charge balance
On the left: \[{2^ - } + {3^ - } + {8^ + } = {3^ + }\]
On the right: \[{6^ + } + {3^ - } = {3^ + }\]
The balanced equation:
\[C{r_2}{O_7}^{2 - } + N{O_2} + 8{H^ + } \to C{r^{3 + }} + N{O_3}^ - + 4{H_2}O\]
Note: Adjusting redox reactions initially requires splitting the condition into the two half-reactions of reduction and oxidation.
All atoms aside from oxygen and hydrogen ought to be balanced first.
In acidic conditions, the oxygen atoms ought to be balanced with water, while hydrogen atoms ought to be balanced with \[{H^ + }\].
In essential conditions, the oxygen atoms ought to be balanced with \[O{H^ - }\], while the hydrogen atoms ought to be balanced with water.
The completely balanced reaction ought to have both half-reactions added back together
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