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sashaice [31]
3 years ago
10

The half-reaction occurring at the anode in the balanced reaction shown below is ________. 3mno4- (aq) + 24h+ (aq) + 5fe (s) → 3

mn2+ (aq) + 5fe3+ (aq) + 12h2o (l)
Chemistry
1 answer:
alexgriva [62]3 years ago
3 0
Reduction takes place in cathode while oxidation takes place at the anode.
Given that the reaction 3MnO4- +24H + +5Fe→3Mn+2+5Fe+3+12H2O
Now that oxidation is termed as an increase in oxidation number or loss of electrons while reduction is a decrease in oxidation number and gain of electrons,
∴oxidation will be
Fe→Fe +3+3e-
reduction will
MnO4- +8H+ →Mn+2+4H2O
If oxidation takes place at anode then Anode: an oxidation reaction
Fe→Fe+3+3e- . Then the answer is 
Fe(s)→Fe+3(aq)+3e-
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Arguments in favor of renewable energy
fenix001 [56]

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5 0
3 years ago
MgO + H2O = Mg(OH)2 is an example of a _____ chemical reaction.
algol [13]

a) synthesis- or combination- reaction

3 0
3 years ago
Calculate the pH at the equivalence point for the titration of 0.110 M methylamine (CH3NH2) with 0.110 M HCl. The Kb of methylam
nikklg [1K]
The reaction between the reactants would be:

CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻

Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.

             CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I                0.11                       0             0
C               -x                          +x           +x
E            0.11 - x                     x             x

Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]

Since the given information is Kb, let's find Ka in terms of Kb.

Ka = Kw/Kb, where Kw = 10⁻¹⁴

So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]

Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>


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jasenka [17]
The reactions of the human body are exothermic
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Please help me and get me the right answer
weqwewe [10]

Answer:

8.21 × 10^{15} Hz

5 0
3 years ago
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