The reaction,K_2Cr_2O_7 + m , FeSO_4 + n , H_ | vedclass

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Question

(B) The given unbalanced equation is: $K_2Cr_2O_7 + m \, FeSO_4 + n \, H_2SO_4 \longrightarrow Cr_2(SO_4)_3 + p \, Fe_2(SO_4)_3 + K_2SO_4 + q \, H_2O$

Vedclass · Accepted Answer

$6, 7, 3, 7$

Vedclass · Answer

(B) The given unbalanced equation is: $K_2Cr_2O_7 + m \, FeSO_4 + n \, H_2SO_4 \longrightarrow Cr_2(SO_4)_3 + p \, Fe_2(SO_4)_3 + K_2SO_4 + q \, H_2O$ Using the oxidation number method or ion-electron method to balance the redox reaction: $1$. Oxidation half-reaction: $Fe^{2+} \longrightarrow Fe^{3+} + e^-$ $2$. Reduction half-reaction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$ Multiplying the oxidation half-reaction by $6$ to balance the electrons: $6Fe^{2+} \longrightarrow 6Fe^{3+} + 6e^-$ Adding the two half-reactions: $K_2Cr_2O_7 + 6FeSO_4 + 7H_2SO_4 \longrightarrow Cr_2(SO_4)_3 + 3Fe_2(SO_4)_3 + K_2SO_4 + 7H_2O$ Comparing the coefficients,we get $m = 6, n = 7, p = 3, q = 7$. Thus,the correct option is $B$.

The reaction,$K_2Cr_2O_7 + m \, FeSO_4 + n \, H_2SO_4 \longrightarrow Cr_2(SO_4)_3 + p \, Fe_2(SO_4)_3 + K_2SO_4 + q \, H_2O$ when balanced,$m, n, p$ and $q$ are,respectively:

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