Redox ReactionsMedium

Oxidation numbers

Determine the oxidation numbers of each element in: H₂SO₄, K₂Cr₂O₇, NaHCO₃.

Given data

Rules: O = -2, H = +1, alkali metals = +1

Steps

0 / 3

In H₂SO₄: n.o.(H)=+1, n.o.(O)=-2. Find n.o.(S) knowing sum is 0.
Write your working — the numerical result must appear in the text
In K₂Cr₂O₇: n.o.(K)=+1, n.o.(O)=-2. Find n.o.(Cr).
In NaHCO₃: n.o.(Na)=+1, n.o.(H)=+1, n.o.(O)=-2. Find n.o.(C).

Full worked solution

In H₂SO₄: n.o.(H)=+1, n.o.(O)=-2. Find n.o.(S) knowing sum is 0.
$2\cdot1 + x + 4\cdot(-2) = 0$
$2\cdot1 + x + 4\cdot(-2) = 0 \rightarrow 2 + x - 8 = 0 \rightarrow x = \mathbf{+6}$ . Sulfur.
In K₂Cr₂O₇: n.o.(K)=+1, n.o.(O)=-2. Find n.o.(Cr).
$2\cdot1 + 2x + 7\cdot(-2) = 0$
$2\cdot1 + 2x + 7\cdot(-2) = 0 \rightarrow 2 + 2x - 14 = 0 \rightarrow x = \mathbf{+6}$ . Chromium.
In NaHCO₃: n.o.(Na)=+1, n.o.(H)=+1, n.o.(O)=-2. Find n.o.(C).
$1+1+x+3\cdot(-2) = 0$
$1+1+x+3\cdot(-2) = 0 \rightarrow 2 + x - 6 = 0 \rightarrow x = \mathbf{+4}$ . Carbon.

Result:H₂SO₄: S = +6 — K₂Cr₂O₇: Cr = +6 — NaHCO₃: C = +4.

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