How do you balance $C r + F e {(N O_{3})}_{2} \to F e + C r {(N O_{3})}_{3}$ $Cr + Fe(NO_3)_2 -> Fe + Cr(NO_3)_3$ ?

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Answer 1 · 2016-12-28T07:58:47

$2 C r + 3 F e {(N O_{3})}_{2} \to 3 F e + 2 C r {(N O_{3})}_{3}$ $2Cr+3Fe(NO_3)_2 rarr 3Fe + 2Cr(NO_3)_3$

We write the oxidation rxn:

$C r \to C r^{3 +} + 3 e^{-}$ $CrrarrCr^(3+) + 3e^-$ $(i)$ $(i)$

And then the reduction rxn:

$F e^{2 +} + 2 e^{-} \to F e$ $Fe^(2+) +2e^(-) rarr Fe$ $(i i)$ $(ii)$

And then cross mulitply, $2 \times (i) + 3 \times (i i)$ $2xx(i) +3xx(ii)$

$2 C r + 3 F e {(N O_{3})}_{2} \to 3 F e + 2 C r {(N O_{3})}_{3}$ $2Cr+3Fe(NO_3)_2 rarr 3Fe + 2Cr(NO_3)_3$

Is mass balanced? Is charge balanced?

How do you balance Cr + Fe(NO_3)_2 -> Fe + Cr(NO_3)_3Cr+Fe(NO3)2→Fe+Cr(NO3)3Cr + Fe(NO_3)_2 -> Fe + Cr(NO_3)_3?