When the redox equation $C r^{3 +} (a q) + 3 M n (s) \to M n^{2 +} (a q) + C r (s)$ $Cr^(3+)(aq) + 3Mn(s) -> Mn^(2+)(aq) + Cr(s)$ is completely balanced, what will the coefficient of $C r^{3 +} (a q)$ $Cr^(3+)(aq)$ be?

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When the redox equation $C r^{3 +} (a q) + 3 M n (s) \to M n^{2 +} (a q) + C r (s)$ $Cr^(3+)(aq) + 3Mn(s) -> Mn^(2+)(aq) + Cr(s)$ is completely balanced, what will the coefficient of $C r^{3 +} (a q)$ $Cr^(3+)(aq)$ be?

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Answer 1 · 2017-06-10T06:35:05

$2...........$

Explanation:

Manganese is oxidized:

$stackrel0Mn(s)rarrMn^(2+) + 2e^(-)$ $(i)$

Chromium is reduced:

$Cr^(3+) + 3e^(-)rarrstackrel0Cr(s)$ $(ii)$

And so we take $3xx(i) + 2xx(ii)$ to eliminate the electrons:

$3stackrel0Mn(s)+2Cr^(3+) rarr3Mn^(2+) + 2stackrel(0)Cr$

Charge is balanced, and mass is balanced, and so this is a reasonable representation of reality. Whether the reaction actually works, I don't know, and cannot be bothered to look up redox tables.

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When the redox equation $C r^{3 +} (a q) + 3 M n (s) \to M n^{2 +} (a q) + C r (s)$ $Cr^(3+)(aq) + 3Mn(s) -> Mn^(2+)(aq) + Cr(s)$ is completely balanced, what will the coefficient of $C r^{3 +} (a q)$ $Cr^(3+)(aq)$ be?

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Explanation:

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When the redox equation Cr^(3+)(aq) + 3Mn(s) -> Mn^(2+)(aq) + Cr(s)Cr3+(aq)+3Mn(s)→Mn2+(aq)+Cr(s)Cr^(3+)(aq) + 3Mn(s) -> Mn^(2+)(aq) + Cr(s) is completely balanced, what will the coefficient of Cr^(3+)(aq)Cr3+(aq)Cr^(3+)(aq) be?

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Explanation:

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When the redox equation $C r^{3 +} (a q) + 3 M n (s) \to M n^{2 +} (a q) + C r (s)$ $Cr^(3+)(aq) + 3Mn(s) -> Mn^(2+)(aq) + Cr(s)$ is completely balanced, what will the coefficient of $C r^{3 +} (a q)$ $Cr^(3+)(aq)$ be?