Question

the redox reaction given below occurs in basic solution.
$\ce{cr_{2}o^{2-}_{7} + i_{2} -> cr^{+3} + io^{-}_{3}}$
balance the half - reaction:
$\ce{cr_{2}o^{2-}_{7} -> cr^{+3}}$
how many electrons are transferred?
$\ce{?e^{-}}$

Explanation:

Step1: Balance Cr atoms

In \( \text{Cr}_2\text{O}_7^{2 -} \), there are 2 Cr atoms, and on the product side, we have \( \text{Cr}^{3 +} \). So we start by balancing Cr:
\( \text{Cr}_2\text{O}_7^{2 -}
ightarrow 2\text{Cr}^{3 +} \)

Step2: Balance O atoms (basic solution, add \( \text{H}_2\text{O} \))

There are 7 O on the left. Add \( 7\text{H}_2\text{O} \) to the right to balance O:
\( \text{Cr}_2\text{O}_7^{2 -}
ightarrow 2\text{Cr}^{3 +} + 7\text{H}_2\text{O} \)

Step3: Balance H atoms (add \( \text{H}^+ \), then convert to basic with \( \text{OH}^- \))

Now, balance H: there are \( 14\text{H} \) in \( 7\text{H}_2\text{O} \), so add \( 14\text{H}^+ \) to the left:
\( \text{Cr}_2\text{O}_7^{2 -} + 14\text{H}^+
ightarrow 2\text{Cr}^{3 +} + 7\text{H}_2\text{O} \)

But since it’s basic, add \( 14\text{OH}^- \) to both sides to neutralize \( \text{H}^+ \) (forming \( \text{H}_2\text{O} \)):
\( \text{Cr}_2\text{O}_7^{2 -} + 14\text{H}^+ + 14\text{OH}^-
ightarrow 2\text{Cr}^{3 +} + 7\text{H}_2\text{O} + 14\text{OH}^- \)
Simplify \( \text{H}^+ + \text{OH}^- = \text{H}_2\text{O} \):
\( \text{Cr}_2\text{O}_7^{2 -} + 14\text{H}_2\text{O}
ightarrow 2\text{Cr}^{3 +} + 7\text{H}_2\text{O} + 14\text{OH}^- \)
Cancel \( 7\text{H}_2\text{O} \) from both sides:
\( \text{Cr}_2\text{O}_7^{2 -} + 7\text{H}_2\text{O}
ightarrow 2\text{Cr}^{3 +} + 14\text{OH}^- \)

Step4: Balance charge (find electrons transferred)

Calculate charges:

• Left: \( \text{Cr}_2\text{O}_7^{2 -} \) has charge \( -2 \), \( 7\text{H}_2\text{O} \) is neutral, total left charge: \( -2 \).
• Right: \( 2\text{Cr}^{3 +} \) has \( 2 \times (+3) = +6 \), \( 14\text{OH}^- \) has \( 14 \times (-1) = -14 \), total right charge: \( +6 - 14 = -8 \).

The charge difference is \( -8 - (-2) = -6 \), meaning we need to add electrons to the left (since left is more positive? Wait, no: charge must balance. Let’s redo charge balance without basic steps (easier for redox):

Original half-reaction (before basic adjustment, using \( \text{H}^+ \)):
\( \text{Cr}_2\text{O}_7^{2 -} + 14\text{H}^+
ightarrow 2\text{Cr}^{3 +} + 7\text{H}_2\text{O} \)

Charge left: \( -2 + 14(+1) = +12 \)
Charge right: \( 2(+3) + 0 = +6 \)

To balance charge, we need to gain electrons (since left is more positive). The difference is \( +12 - +6 = +6 \), so we add \( 6e^- \) to the left (reduction, gain electrons):

\( \text{Cr}_2\text{O}_7^{2 -} + 14\text{H}^+ + 6e^-
ightarrow 2\text{Cr}^{3 +} + 7\text{H}_2\text{O} \)

Check charge: left \( -2 +14 -6 = +6 \), right \( +6 +0 = +6 \). Balanced.

So the number of electrons transferred (gained, in this reduction half-reaction) is 6.

Answer:

6