Question

pleae solve all steps in problems 3 & 4

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Answer 1

31 Given Unbalanced Reaction a) Mnoata >mnttel (inaccdic medium) mng tuit one dations Half reaction Reduchess Halfleachon ad ch Mno Mn2ttato zu ehtze - mm2 +2160 Mngfultttze- met +2120-11 It it 2t htzé Mno, tuft the sam24+240 + 2H2O imand "Mno tree tutt mn't tell 2 24 mng Mną tel mn2 + c (in basic median oxidation Half reaction Reduction Huf reaction auch tato zat schtré -I Mno tylko talbetroH Mng tuhoté mont takot 40147 mng mnt 24 um Pt II.3 20 cht aé -I - Mnatubotze 2t s mri +21201 404 Mn2t tel +2160 tuolf ung tret +4150 en Basic medium

I + 360 -12 +3120 Gren Unbalanced quation, 4.503 + 103 sop2- +I2 can acidic median) oxidation Hulf reation Reduction Halfeeahon Itsoz Sap 2 Itsuz tho sou 2- I 03-12 102 → 1853-t60 Sq 27347 103 76itt Itsa tho soy/ +31ttze 103.76#*+5é > Izt340 -I -IT I X5+ D X 2 Sits og & sito 5 Sey2 +15Httloé 283 - tlaltt toe 212t 640 on acidic 5-H 503 +21037 5s02-+25+ 3tt +20 its of tiz--s soy2 +52 ( in Basic medium) Reduction Half reachen o udahlo Kaffeeautim, Ia Iz If50% - sap 2- 103 lat 3 bo Its of tho sau 7 3160 I 03 +6 to → Izt3140760H issotho+30H Asos + 10 +3014"-350 7310426 10307614075€ 14+3160 -I toolf ssay 2- -T Inst 102 54s8+510H150H55002- +15 hotive 2803 +12160 + 10e 2 Int 6160 + 12014 51:593 +213 +30476 43 5572 F222 15 to +229 +308 761420 5 soy2 +20 en Basite me ediam.
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