1868 United States presidential election in Iowa

1868 United States presidential election in Iowa

← 1864 November 3, 1868 1872 →
 
Nominee Ulysses S. Grant Horatio Seymour
Party Republican Democratic
Home state Illinois New York
Running mate Schuyler Colfax Francis Preston Blair, Jr.
Electoral vote 8 0
Popular vote 120,399 74,040
Percentage 61.92% 38.08%

County Results

Grant

  50-60%
  60-70%
  70-80%
  80-90%
  90-100%

Seymour

  50-60%
  60-70%

Tie

  ~50%


President before election

Andrew Johnson
Democratic

Elected President

Ulysses S. Grant
Republican

Elections in Iowa
U.S. Presidential elections
Presidential caucuses
Democratic
1976
1980
1984
1992
2000
2004
2008
2012
2016
2020
2024
Republican
1980
1996
2000
2008
2012
2016
2020
2024
U.S. Senate elections
U.S. House of Representatives elections
General elections
Gubernatorial elections
Secretary of State elections
Attorney General elections
Senate elections
House of Representatives elections
Territorial Council elections
Ballot measures
Mayoral elections
  • v
  • t
  • e

The 1868 United States presidential election in Iowa took place on November 3, 1868, as part of the 1868 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for president and vice president.

Iowa voted for the Republican nominee, Ulysses S. Grant, over the Democratic nominee, Horatio Seymour. Grant won the state by a margin of 23.84%.

Results

1868 United States presidential election in Iowa[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Ulysses S. Grant of Illinois Schuyler Colfax of Indiana 120,399 61.92% 8 100.00%
Democratic Horatio Seymour of New York Francis Preston Blair, Jr. of Missouri 74,040 38.08% 0 0.00%
Total 194,439 100.00% 8 100.00%

See also

References

  1. ^ "1868 Presidential General Election Results - Iowa".


Stub icon 1

This Iowa elections-related article is a stub. You can help Wikipedia by expanding it.

  • v
  • t
  • e