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Redox Reactions question

2020 · 2 Sep · Shift 2 · Q4

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JEE Main
Chemistry
Redox Reactions
2020 · 2 Sep · Shift 2 · Q4

Redox Reactions question

2020 · 2 Sep · Shift 2 · Q4

JEE MainChemistryRedox ReactionsNumerical+4 / −1

The oxidation states of transition metal atoms in

K_2Cr_2O_7

,

KMnO_4

and

K_2FeO_4

, respectively, are x, y and z. The sum of x, y and z is

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 19

Find oxidation state of Cr in $K_2Cr_2O_7$

Let the oxidation state of Cr be $x$ .

Using charge balance: $2(+1) + 2x + 7(-2) = 0$ $2 + 2x - 14 = 0$ $2x - 12 = 0$ $2x = 12$ $x = +6$

Find oxidation state of Mn in $KMnO_4$

Let the oxidation state of Mn be $y$ .

Using charge balance: $+1 + y + 4(-2) = 0$ $1 + y - 8 = 0$ $y - 7 = 0$ $y = +7$

Find oxidation state of Fe in $K_2FeO_4$

Let the oxidation state of Fe be $z$ .

Using charge balance: $2(+1) + z + 4(-2) = 0$ $2 + z - 8 = 0$ $z - 6 = 0$ $z = +6$

Sum of oxidation states

$x + y + z = 6 + 7 + 6 = 19$

Therefore, the required integer is: $\boxed{19}$

Comparison with stored correct answer

Stored correct answer = $19$

Our derived answer also equals $19$ , so the answers agree.

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