| Election Name: | 1875 Iowa Senate election |
| Country: | Iowa |
| Flag Image: | Flag of Iowa (variant).svg |
| Type: | legislative |
| Ongoing: | no |
| Previous Election: | 1873 Iowa Senate election |
| Previous Year: | 1873 |
| Next Election: | 1877 Iowa Senate election |
| Next Year: | 1877 |
| Seats For Election: | 30 out of 50 seats in the Iowa State Senate |
| Majority Seats: | 26 |
| Election Date: | October 12, 1875 |
| Party1: | Republican Party (United States) |
| Last Election1: | 36 |
| Seat Change1: | 4 |
| Seats After1: | 40 |
| Party2: | Democratic Party (United States) |
| Last Election2: | 10 |
| Seats After2: | 10 |
| Party3: | Anti-Monopoly Party (US) |
| Last Election3: | 4 |
| Seat Change3: | 4 |
| Seats After3: | 0 |
In the 1875 Iowa State Senate elections, Iowa voters elected state senators to serve in the sixteenth Iowa General Assembly. Elections were held in 30 of the state senate's 50 districts. State senators serve four-year terms in the Iowa State Senate.
The general election took place on October 12, 1875.[1]
Following the previous election, Republicans had control of the Iowa Senate with 36 seats to Democrats' 10 seats and four members of the Anti-Monopoly Party.
To claim control of the chamber from Republicans, the Democrats needed to net 16 Senate seats.
Republicans maintained control of the Iowa State Senate following the 1875 general election with the balance of power shifting to Republicans holding 40 seats and Democrats having 10 seats (a net gain of 4 seats for Republicans).
District boundaries for the Iowa Senate in 1875: