1868 United States presidential election in Iowa

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.

1868 United States presidential election in Iowa

← 1864 November 3, 1868 1872 →
  Ulysses S Grant by Brady c1870-restored (cropped).jpg Horatio Seymour - Brady-Handysmall.jpg
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%

Iowa Presidential Election Results 1868.svg
County Results

President before election

Andrew Johnson

Elected President

Ulysses S. Grant

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%.


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 alsoEdit


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