1864 United States presidential election in Iowa

Last updated

1864 United States presidential election in Iowa
Flag of Iowa.svg
  1860 November 8, 1864 1868  
Turnout19.70% of the total population Increase2.svg 0.63 pp [1]
  Abraham Lincoln November 1863.jpg GeorgeMcClellan2 (cropped).jpg
Nominee Abraham Lincoln George B. McClellan
Party National Union Democratic
Home state Illinois New Jersey
Running mate Andrew Johnson George H. Pendleton
Electoral vote80
Popular vote88,50049,525
Percentage64.12%35.88%

Iowa Presidential Election Results 1864.svg
County Results

President before election

Abraham Lincoln
Republican

Elected President

Abraham Lincoln
National Union

The 1864 United States presidential election in Iowa took place on November 8, 1864, as part of the 1864 United States presidential election. Iowa voters chose eight representatives, or electors, to the Electoral College, who voted for president and vice president. [2]

Contents

Iowa was won by the National Union candidate Republican incumbent President Abraham Lincoln of Illinois and his running mate former Senator and Military Governor of Tennessee Andrew Johnson. They defeated the Democratic candidate 4th Commanding General of the United States Army George B. McClellan of New Jersey and his running mate Representative George H. Pendleton of Ohio. Lincoln won the state by a margin of 28.24%. [2]

Results

1864 United States presidential election in Iowa [2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
National Union Abraham Lincoln of Illinois Andrew Johnson of Tennessee 88,50064.12%8100.00%
Democratic George B. McClellan of New Jersey George H. Pendleton of Ohio 49,52535.88%00.00%
Total138,025100.00%8100.00%

See also

References

  1. "1880 Presidential Election Results Iowa Total Population Turnout".
  2. 1 2 3 "1864 Presidential Election Results Iowa".