Question

Balance the following redox equations by the half-reaction method:

(a) Mn²⁺ + H₂O₂ → MnO₂ + H₂O (in basic solution)

(b) Bi(OH)₃ + SnO₂²⁻ → SnO₃²⁻ + Bi (in basic solution)

(c) Cr₂O₇²⁻ + C₂O₄²⁻ → Cr³⁺ + CO₂ (in acidic solution)

(d) ClO₃⁻ + Cl⁻ → Cl₂ + ClO₂ (in acidic solution)

(e) Mn²⁺ + BiO₃⁻ → Bi³⁺ + MnO₄⁻ (in acidic solution)

Answer 1

(a) 2+ solution. mot tho u hoe & tho in basic Reducing agent oxidising agent in How we Convert ting into two halt equation bor oxidation and another tor reduction. one 2 + Muyot 1 mno & 2140 +2e. an 20 at the the tho +2014 JT. Met & tho2+2014 uno + 2tho 14 In this method first atom of balanced, then che balanced by adding electron to make eren lide neretral. Then electrons in each halb equation should be Cacuated by multiplying whole number in the half equations. Then two halb canations are added to get balanced (b) +3 a e equation. 12 - & BioH Snor Snoz - - no 2 all t Bi basic solution) oxiditing agent Reduling agent Bilohi, tze shoz ² tgon. Bi 4 301 JX2 shoz?- + the the ] x3 - 2 BiGH & 3 snor i Bi + 3500 + 3 the

(C) #z Croz Cy+3+ to 34²- Reduling , acidic medium, oxidising agent agent 2- Eroz + 14 H7+60 2873 + 7420 2- you Note on acidic medi hydrogen att y adde – 2002 the ] x3 6607 +1441 +2602- 26873 + 600+ 740 alidic medium to balance in H0 is added h to that lide oxygen is deficient ince and to ! i geroogen, a tt y added . Then charge balanced by addition of electron. Then electron loss and gain. should be equal in halb equation by multiplying whole number then the two halls equations are added. to hol & cut chit thon (inalidic medians) (oxiditing agent) (Reducing agent) con + 2H+ te co2 + 140 142 a zat cl the + 2tho - 2012 + c zelo tacê tuft medium & myat the Broth Bit & nos falidir (Reducing agent, oxiditing agent Mn21441450 - innan +8H+ +5eJX2 3:01 +6Ht the Bit 3 + 3 the j x5 2 Mn't + 5 Biol- & 14 H a mna + 5 Bit3 +7tho