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1864 United States presidential election in Iowa

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1864 United States presidential election in Iowa

← 1860November 8, 18641868 →
Turnout19.70% of the total populationIncrease 0.63pp[1]
 
NomineeAbraham LincolnGeorge B. McClellan
PartyNational UnionDemocratic
Home stateIllinoisNew Jersey
Running mateAndrew JohnsonGeorge H. Pendleton
Electoral vote80
Popular vote88,50049,525
Percentage64.12%35.88%

County Results

Lincoln

  50–60%
  60–70%
  70–80%
  80–90%
  90–100%

McClellan

  50–60%
  60–70%
  70–80%
  80–90%

Tie

  ~50%


President before election

Abraham Lincoln
Republican

Elected President

Abraham Lincoln
National Union

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The1864 United States presidential election in Iowa took place on November 8, 1864, as part of the1864 United States presidential election. Iowa voters chose eight representatives, or electors, to theElectoral College, who voted forpresident andvice president.[2]

Iowa was won by theNational Union candidateRepublican incumbentPresidentAbraham Lincoln ofIllinois and his running mate formerSenator andMilitary Governor of TennesseeAndrew Johnson. They defeated theDemocratic candidate 4thCommanding General of the United States ArmyGeorge B. McClellan ofNew Jersey and his running mateRepresentativeGeorge H. Pendleton ofOhio. Lincoln won the state by a margin of 28.24%.[2]

Results

[edit]
1864 United States presidential election in Iowa[2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
National UnionAbraham Lincoln ofIllinoisAndrew Johnson ofTennessee88,50064.12%8100.00%
DemocraticGeorge B. McClellan ofNew JerseyGeorge H. Pendleton ofOhio49,52535.88%00.00%
Total138,025100.00%8100.00%

See also

[edit]

References

[edit]
  1. ^"1880 Presidential Election Results Iowa Total Population Turnout".
  2. ^abc"1864 Presidential Election Results Iowa".
Electoral map, 1864 election


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